Let S_k, k=1,2, , 100 , denote the sum of the infinite geometric series whose first term is (k-1)/(k !) and the common ratio is (1)/(k) . Then the value of (100^2)/(100 !)+∑_ k=1 ^ 100 |(k^2-3 k+1) S_k| is

Correct answer: 0 We have, S_k=((k-1)/(k !))/(1-(1)/(k))=(1)/((k-1) !) Now, aligned & (k^2-3 k+1) S_k=\ (k-2)(k-1)-1\ & =(1)/((k-3) !)-(1)/((k-1) !) & ∑_ k=1 ^ 100 |(k^2-3 k+1) S_k| & =1+1+2-((1)/(99 !)+(1)/(98 !))=4-(100^2)/(100 !) aligned (100^2)/(100 !)+∑_ k=1 ^ 100 |(k^2-3 k+1) S_k|=4

JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Sequences and Series

JEE Advanced 2010 · Paper 1 · Mathematics · Sequences and Series

Let S_k, k=1,2, , 100 , denote the sum of the infinite geometric...

Let

S_k, k=1,2, \ldots, 100

, denote the sum of the infinite geometric series whose first term is

\frac{k-1}{k !}

and the common ratio is

\frac{1}{k}

. Then the value of

\frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|

Correct answer

Step-by-step solution

We have,

S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}

Now,

\begin{aligned}& \left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \\ & =\frac{1}{(k-3) !}-\frac{1}{(k-1) !}\\ & \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \\ & =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !} \end{aligned}

\Rightarrow \quad \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4

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